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17 April, 20:20

At noon, ship a is 60 km west of ship

b. ship a is sailing south at 15 km/h and ship b is sailing north at 5 km/h. how fast is the distance between the ships changing at 4:00 pm?

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  1. 17 April, 20:41
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    DA/dt=15 km/h and dB/dt=5 km/h

    base of our triangle if drawn is 60 km

    to find dD/dt when t=4 hours we shall have:

    (A+B) ²+60²=D²

    d/dt[ (A+B) ²+60²]=d/dt (D²)

    [d (A+B) ]/dt*2 (A+B) (2) = dD/dt*2D

    B=5km/h*4hr=20km

    A=15km/hr*4=60km

    20²+60²=D²

    D=√4000

    (15+5) * (2) * 80 * (2) = 2*√4000*dD/dt

    dD/dt=6400/√4000=101.192 km/hr
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